In \(\triangle RUV\)
\(RV^2 = 4^2 + 2^2\)
\(RV = 2\sqrt{5}\)

In \(\triangle RSW\)
\(RW^2 = 4^2 + 2^2\)
\(RW = 2\sqrt{5}\)

In \(\triangle WTV\)
\(WV^2 = 2^2 + 2^2\)
\(WV = 2\sqrt{2}\)

Now, we can see that \(\triangle RWV\) is an isosceles triangle. We can apply either Heron's formula or drop a perpendicular from point R to base VW (at X)
\((2\sqrt{5})^2 = RX^2 + (\sqrt{2})^2\)
\(RX = \sqrt{(20 - 2)} = \sqrt{18} = 3\sqrt{2}\)

Area of \(\triangle RWV\) = \(\frac{1}{2}(2\sqrt{2})(3\sqrt{2}) = \frac{1}{2}(12) = 6\)

Hence, option B